链接:https://leetcode-cn.com/problems/delete-and-earn/solution/javadong-tai-gui-hua-100da-jia-jie-she-b-n5o7/
解题思路
没做过打家劫舍的可以先做做198.打家劫舍
创建一个新数组count,对nums[i]计数,将这个题转换为打家劫舍的思路
推导公式为 dp[i] = Math.max(dp[i-1], dp[i-2]+count[i]*i);
java代码:
class Solution { public int deleteAndEarn(int[] nums) { int n = nums.length; if(n <= 1 ) return nums[0]; int max = nums[0]; for(int i : nums) { max = Math.max(max, i); } int[] count = new int[max+1]; for(int i : nums) { count[i]++; } int[] dp = new int[max+1]; dp[1] = count[1]*1; dp[2] = Math.max(dp[1], count[2]*2); for(int i = 3; i <= max; i++) { dp[i] = Math.max(dp[i-1], dp[i-2]+count[i]*i); } return dp[max]; } }
c++代码:
class Solution { public: int deleteAndEarn(vector<int>& nums) { int n = nums.size(); if( n == 1) return nums[0]; int maxnum = nums[0]; for(int i : nums) { maxnum = max(i,maxnum); } vector<int> count(maxnum+1),dp(maxnum+1); for(int i : nums) { count[i]++; } dp[1] = count[1]; for(int i = 2; i <= maxnum; i++) { dp[i] = max(dp[i-1],dp[i-2] + count[i]*i); } return dp[maxnum]; } };